C - Arrangement for Contests

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题目链接
维护一个区间最小值同时维护一个区间最小值的减法

#include <bits/stdc++.h>

using namespace std;
const int N = 100010;
int a[N];
int n,k;
struct node{
    int l,r;
    int v,add;
}tr[N * 4];

void pushup(int u)
{
    tr[u].v = min(tr[u << 1].v , tr[u << 1 | 1].v);
}

void pushdown(int u)
{
    auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if(root.add){
        left.add += root.add, left.v -= root.add;
        right.add += root.add, right.v -= root.add;
        root.add = 0;
    }
}

void build(int u,int l,int r)
{
    if(l == r) tr[u] = {r,r,a[r],0};
    else{
        tr[u] = {l,r};
        int mid = l + r >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u,int l,int r,int d)
{
    if(tr[u].l == l && tr[u].r == r){
        tr[u].v -= d;
        tr[u].add += d;
    }
    else{
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if(r <= mid) modify(u << 1, l, r, d);
        else if(l > mid) modify(u << 1 | 1, l , r ,d);
        else modify(u << 1, l , mid,d), modify(u << 1 | 1, mid + 1, r, d);
        pushup(u);
    }
}

int query(int u,int l,int r)
{
    if(tr[u].l == l && tr[u].r == r) return tr[u].v;

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if(r <= mid) return query(u << 1,l , r);
    else if(l > mid) return query(u << 1 | 1, l,  r);
    else return min(query(u << 1, l , mid), query(u << 1 | 1, mid + 1, r));
}

signed main()
{
    int _; cin >> _;
    while(_ --){
        scanf("%lld%lld",&n,&k);
        for(int i = 1; i <= n; i ++) scanf("%lld",&a[i]);
        build(1,1,n);
        int res = 0;
        for(int i = k; i <= n; i ++){
            int t = query(1,i - k + 1,i);
            res += t;
            modify(1,i - k + 1,i, t);
        }
        cout << res << endl;
    }
}

C - Arrangement for Contests

标签：else   题目   cout   build   root   bit   lse   names   res   

原文地址：https://www.cnblogs.com/acm-Patrick/p/14725874.html